91Ó°ÊÓ

Let and be piecewise continuous on $[0,\mathrm{∞})$and of exponential orderand set,$\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathcal{L}\mathbf{\left\{}\mathit{f}\mathbf{\right\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\text{and}}\mathit{G}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathcal{L}\mathbf{\left\{}\mathit{g}\mathbf{\right\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$ then,

$\mathcal{L}\mathbf{\{}\mathit{f}\mathbf{∗}\mathit{g}\mathbf{\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathit{G}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{,}$or

${\mathcal{L}}^{−1}\left\{F\right(s\left)G\right(s\left)\right\}\left(t\right)=(f∗g)\left(t\right)$

Consider the given equation,

$\frac{s}{(s−1)(s+2)}$

Let,

$\begin{array}{c}y\left(s\right)=\frac{s}{(s−1)(s+2)}\\ =\frac{1}{s+2}[1+\frac{1}{s−1}]\\ =\frac{1}{s+2}+\frac{1}{(s−1)(s+2)}\end{array}$

Take inverse Laplace transform on both sides,

${\mathcal{L}}^{−1}\left[y\right(s\left)\right]={\mathcal{L}}^{−1}\left[\frac{{\displaystyle 1}}{{\displaystyle s+2}}\right]+{\mathcal{L}}^{−1}\left[\frac{{\displaystyle 1}}{{\displaystyle (s−1)(s+2)}}\right]$$→\left(1\right)$

Hence, the convolution formula is,${\mathcal{L}}^{−1}\left[f\right(s)â‹\dots g(s\left)\right]=f⋆g={∫}_0^{t}f(t−v)g\left(v\right)dv$, where

$f\left(s\right)=\frac{1}{s−1}$and$f\left(t\right)=e^t$

$g\left(s\right)=\frac{1}{s+2}$and $g\left(t\right)=e^{−2t}$

Thus, the equationcan be written as,

$\begin{array}{c}y\left(t\right)=e^{−2t}+{∫}_0^t{e}^{t−v}â‹\dots e^{−2v}dv\\ =e^{−2t}+e^t{∫}_0^t{e}^{−3v}dv\\ =e^{−2t}+e^t{\left[\frac{e^{−3v}}{−3}\right]}_0^t\\ =e^{−2t}−\frac{e^t}{3}[e^{−3t}−1]\end{array}$

$\begin{array}{c}y\left(t\right)=e^{−2t}−\frac{e^{−2t}}{3}+\frac{e^t}{3}\\ =\frac{2e^{−2t}}{3}+\frac{e^t}{3}\end{array}$

Therefore, the inverse Laplace transform for the given function is.$y\left(t\right)=\frac{2e^{−2t}}{3}+\frac{e^t}{3}$